Problem: Simplify and expand the following expression: $ \dfrac{4}{z + 1}+ \dfrac{2}{z + 6}+ \dfrac{3z}{z^2 + 7z + 6} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{3z}{z^2 + 7z + 6} = \dfrac{3z}{(z + 1)(z + 6)}$ Now we have: $ \dfrac{4}{z + 1}+ \dfrac{2}{z + 6}+ \dfrac{3z}{(z + 1)(z + 6)} $ The least common multiple of the denominators is: $ (z + 1)(z + 6)$ In order to get the first term over $(z + 1)(z + 6)$ , multiply by $\dfrac{z + 6}{z + 6}$ $ \dfrac{4}{z + 1} \times \dfrac{z + 6}{z + 6} = \dfrac{4(z + 6)}{(z + 1)(z + 6)} $ In order to get the second term over $(z + 1)(z + 6)$ , multiply by $\dfrac{z + 1}{z + 1}$ $ \dfrac{2}{z + 6} \times \dfrac{z + 1}{z + 1} = \dfrac{2(z + 1)}{(z + 1)(z + 6)} $ Now we have: $ \dfrac{4(z + 6)}{(z + 1)(z + 6)} + \dfrac{2(z + 1)}{(z + 1)(z + 6)} + \dfrac{3z}{(z + 1)(z + 6)} $ $ = \dfrac{ 4(z + 6) + 2(z + 1) + 3z} {(z + 1)(z + 6)} $ Expand: $ = \dfrac{4z + 24 + 2z + 2 + 3z}{z^2 + 7z + 6} $ $ = \dfrac{9z + 26}{z^2 + 7z + 6}$